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If G is a finite group containing an irreducible character of degree 5, deduce its order.

I know that the order of G is divisble by 5

I also know that the order of G is equal to the sum of the squares of the degrees of irreducible characters.

However I am unable to deduce the order of G with just this information.

Could someone please please please help me determine the exact order of G.

Many thanks Sarah

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If $G$ is a group with an irrep. of order five and $G'$ is any group, then $G \times G'$ has an irrep. of order five (namely, the tensor product of $G$'s irrep. and the trivial rep of $G'$). So the order of $G$ is not determined with this information. –  Akhil Mathew Oct 11 '10 at 22:20
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Are you sure this is all the information you have? –  Qiaochu Yuan Oct 12 '10 at 20:05
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2 Answers 2

up vote 6 down vote accepted

I am going to assume that you meant to state that this was a faithful representation, meaning that the map $G \to GL_n(\mathbb{C})$ has no kernel. Otherwise, as Akhil says, the problem is dramatically underconstrained.

An infinite class of examples is the semidriect product of $S_5$ acting on $(\mathbb{Z}/N)^4$, for $N$ relatively prime to $30$. In other words, look at all permutation matrices whose nonzero entries are $N$-th roots of unity, with product $1$.

Here is another infinite family of examples: Let $G$ be any subgroup of $GL_5(\mathbb{C})$. Let $A$ be an abelian cyclic group, embedded in $GL_5(\mathbb{C})$ as a subgroup of the scalar matrices. Then $A$ is central in $GL_5(\mathbb{C})$, so $GA$ is a finite subgroup of $GL_5(\mathbb{C})$. If $\mathbb{C}^5$ is an irreducible representation of $G$, it is also an irreducible representation of $GA$.

Roughly speaking, the rest of my answer says that every example looks like one of these.


There is a theorem of Jordan which states that there is a bound $J(n)$ such that any subgroup $G$ of $GL_n(\mathbb{C})$ has a normal abelian subgroup $A$ with $|G/A| \leq J(n)$. See Terry Tao's blog for a nice exposition.

Using modern improvements on Jordan's result, we can get some more precise bounds.

Specifically, every $G$ as in your question is either:

(1) An extension $0 \to A \to G \to H \to 0$, where $A$ is abelian and $H$ is a subgroup of $S_5$ or

(2) A central extension $0 \to A \to G \to H \to 0$ where $|H| \leq 25920$ and $A$ is cyclic.

I think if I knew more about group cohomology I could, in principle, turn this into an infinite but comprehensible, list of all groups that may occur.


A representation $V$ of $G$ is called primitive it is irreducible and not induced from a proper subgroup $H$ of $G$.

Case 1: $\mathbb{C}^5$ is an irreducible, but imprimitive representation, of $G$. Let $\mathbb{C}^5$ be induced from a representation $L$ of $B \subset G$. Then $(\dim L) |G/B| = 5$, so $L$ is one dimensional and $H$ has index $5$ in $G$. Let $A = \bigcap_{g \in G/B} g B g^{-1}$. So $A$ is normal and $G/A$ acts faithfully on the $5$ element set $G/B$, showing that $G/A$ is a subgroup of $S_5$. We must show that $A$ is abelian. Suppose not, and let $C$ be the commutator of $A$. Then $C$ is killed by the map to $GL(L)$, as $L$ is one dimensional. But $C$ is normal in $G$, so it is also killed by the map to $GL(\mathbb{C}^5)$. This contradicts that we are supposed to be dealing with a faithful representation.

Case 2: $\mathbb{C}^5$ is primitive. According to a recent paper of Collins, if $G$ has a a primitive representation of dimension $5$, then $|G/Z(G)| \leq 25920$. (Here $Z(G)$ is the center of $G$.) We take $A=Z(G)$. Since $\mathbb{C}^5$ is irreducible, we know that $Z(G)$ acts by scalars. So $Z(G)$ is a subgroup of $\mathbb{C}^*$, and must be cyclic.

I couldn't find a list of all groups that could occur as $G/Z(G)$, but perhaps you can extract it from Collins' work.

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Thanks, I think the question WAS missing something. After many many emails to my lecturer (I'm studying at a distance learning university) he replied stating I must find a lower bound for G. Thanks again to all the help! :) –  sarah jamal Oct 13 '10 at 3:54
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Although it seems to be well beyond what was intended by Sarah's lecturer, let me point out that the structure of the finite irreducible subgroups of ${\rm GL}(5,\mathbb{C})$ has been well understood since the mid-1960s. Let $G$ be one such. As David Speyer point out, the case of imprimitive groups of degree 5 is pretty transparent, so that leaves the case of primitive groups. These were classified by Richard Brauer around 1965. We can replace the generators of $G$ by scalar multiples and assume that $G \subseteq {\rm SL}(5,\mathbb{C})$. This does not affect the structure of $G/Z(G)$, which is as much as you can hope to control in any case. The problem then splits into two cases: if $G$ has a non-Abelian Sylow $5$-subgroup, then $G$ has a normal $5$-subgroup $U$ of order $125$ such that $G/U$ is isomorphic to a subgroup of ${\rm SL}(2,5)$, and the case $G/U \cong {\rm SL}(2,5)$ does occur. If $G$ has an Abelian Sylow $5$-subgroup, then $Z(G)$ is a direct factor of $G$ of order dividing $5,$ and we may as well suppose that $Z(G) = 1$. The list of possibilities for $G$ is then ${\rm PSL}(2,11)$, ${\rm PSU}(4,2)$, $A_{6}$ or $S_{6}$ (Brauer's list contains some extra (imprimitive) groups, which could have been omitted, ie $A_5$ and $S_5$). My reason for pointing this out is that this sort of problem soon gets very difficult (the classification of finite irreducible subgroups of ${\rm GL}(4,\mathbb{C})$ was done in the relatively early 20-th century, so there had been a long hiatus before $5$-dimensional groups were dealt with). The work of Collins, and Weisfeiler before him, uses the full strength of the classification of the finite simple groups. I also would like to point out that a number of students of Brauer and Walter Feit, and other mathematicians too, amassed a good deal of information about finite complex linear groups in low dimensions from the mid-60s through the early 1980s, when the classification of finite simple groups changed things. To answer the question that Sarah's lecturer intended to ask, it looks as though no finite group $G$ of order less than $55$ has a complex irreducible character of degree $5$. While this needs nothing like the full classification stated here, it does not seem entirely obvious. Depending what background the students had, this seems quite tricky.

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